Question

(solve manually) a randomn sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes.

a) Construct a 95% confidence interval for the true average amount of time customers spent in the restaurant.

b) with a .95 probability, how large of a sample would have to be taken to provide a margin of error of 2.5 minutes or less?

Answer 1

sample size, n = 49

average time, = 45

standard deviation, = 14

degree of freedom, df = 49 - 1 = 48

alpha = 0.05( since 95% confidence interval is given)

a.) Confidence inerval is given as:

   z * ( / )

z (alpha = 0.05) = 1.96

so, putting values we get,

45 1.96 * ( 14 / 7)

45   3.92

so, CI = (41.08, 48.92)

b.) to find a sample so that our error should be less than or equal to 2.5

the formula is:

n = [( z * ) / E] ²

where E = Expected error = 2.5

So, n = (1.96 * 14)² / 6.25

n = (3.8416 * 196) / 6.25

n = 120.47

approximately, n = 120